Use Green's Theorem to evaluate the line integral along the given positively oriented curve. ∫C (3y + 7e^sqrt(x)) dx + (8x + 5 cos y^2) dy C is the boundary of the region enclosed by the parabolas y = x2 and x = y2

∫c ((3y + 7e^(√x) dx + (8x + 5 cos (y²)) dy) = ∫∫ₐ 5dA =  5/3

Step-by-step explanation:

Given that;

∫c ((3y + 7e^(√x) dx + (8x + 5 cos (y²)) dy)

Green's Theorem is given as;

∫c (P(x,y)dx + Q(x,y)dy) = ∫∫ₐ { (-β/βy) P(x,y) + (β/βy) Q(x,y) } dA

Now our P(x,y) = 3y + 7e^(√x) and our Q(x,y) = 8x + 5 cos (y²)

Since we know this, therefore; we substitute

∫c ((3y + 7e^(√x) dx + (8x + 5 cos (y²)) dy) = ∫∫ₐ { (-β/βy) (3y + 7e^(√x))  + (β/βy) (8x + 5 cos (y²)) } dA

∫c ((3y + 7e^(√x) dx + (8x + 5 cos (y²)) dy) = ∫∫ₐ ( 8-3) dA

∫c ((3y + 7e^(√x) dx + (8x + 5 cos (y²)) dy)  = ∫∫ₐ 5dA

from the question, our region is defined by a lower bound: y = x² and an upper bound of y = √x

going from x = 0 to x = 1

Now calculating ∫∫ₐ 5dA  by means of the description of the region, we say;

∫∫ₐ 5dA  = 5¹∫₀   ₓ²∫^(√x) dydx

∫∫ₐ 5dA =  5¹∫₀ (y)∧(y-√x) ∨(y-x²)  dx

∫∫ₐ 5dA = 5¹∫₀ (√x-x²) dx

∫∫ₐ 5dA = 5 [ ((x^(3/2))/(3/2)) - x³/3]¹₀     NOW since ∫[f(x)]ⁿ dx = ([f(x)]ⁿ⁺¹ / n+1) + C

then

∫∫ₐ 5dA = 5 [ ((1^(3/2))/(3/2)) - 1³ / 3) - ((0^(3/2))/(3/2)) - 0³ / 3) ]

∫∫ₐ 5dA = 5 [ ((1^(3/2))/(3/2)) - 1³ / 3)

∫∫ₐ 5dA = 5/3

Therefore ∫c ((3y + 7e^(√x) dx + (8x + 5 cos (y²)) dy) = ∫∫ₐ 5dA =  5/3

hi there, i would be interested to you with your test, may you add the questions and answer options.

- : )

210/22 = 9.55

since he can't take part of a person to the game the answer is 9 people

hope this !